NCERT Solutions for Exercise 3.3 Class 11 Maths Chapter 3

NCERT Solutions for Exercise 3.3 Class 11 Maths Chapter 3 - Trigonometric Functions

Edited By Sumit Saini | Updated on Jul 29, 2022 03:19 PM IST

Questions based on different concepts are asked in Exercise 3.3 Class 11 Maths which includes proof related questions of trigonometric functions. NCERT Solutions for class 11 maths chapter 3 exercise 3.3 is relatively difficult as compared to previous exercises. These questions are asked only in subjective papers like School final examinations. Exercise 3.3 Class 11 Maths questions might take more time to solve but once done, one can expect a good score in the examination. NCERT Solutions for class 11 maths chapter 3 exercise 3.3 is a must to do exercise. Also for the other exercises of NCERT one can find the below.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

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NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

Question:1 Prove that $\small \sin ^{2} \left ( \frac{\pi }{6} \right ) + \cos ^{2}\left ( \frac{\pi }{3} \right ) - \tan ^{2}\left ( \frac{\pi }{4} \right ) = -\frac{1}{2}$

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

$\sin \left ( \frac{\pi}{6} \right ) = \left ( \frac{1}{2} \right )\\ \\ \cos \left ( \frac{\pi}{3} \right ) = \left ( \frac{1}{2} \right )\\ \\ \tan \left ( \frac{\pi}{4} \right ) = 1$
$\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=$ $\left ( \frac{1}{2} \right )^{2}+ \left ( \frac {1}{2} \right )^{2}-1^{2}$

$= \frac{1}{4}+\frac{1}{4}-1= -\frac{1}{2}$
= R.H.S.

Question:2 Prove that $\small 2\sin ^{2}\left ( \frac{\pi }{6} \right ) + cosec ^{2}\left ( \frac{7\pi }{6} \right )\cos ^{2}\frac{\pi }{3} = \frac{3}{2}$

Answer:

The solutions for the given problem is done as follows.

$\sin\frac{\pi}{6} = \frac {1}{2}\\ \\ cosec\frac{7\pi}{6} = cosec\left ( \pi + \frac{\pi}{6} \right ) = -cosec \frac{\pi}{6}=-2\\ \\ \cos \frac{\pi}{3} = \frac{1}{2}$
$2\sin^{2}\frac{\pi}{6} +cosec^{2}\frac{7\pi}{6}\cos^{2}\frac{\pi}{3} = 2\left ( \frac{1}{2} \right )^{2}+\left ( -2 \right )^{2}\left ( \frac{1}{2} \right )^{2}\\ \\ \Rightarrow 2\times\frac{1}{4} + 4\times\frac{1}{4} = \frac {1}{2} + 1= \frac{3}{2}$
R.H.S.

Question:3 Prove that $\small \cot ^{2}\left ( \frac{\pi }{6} \right ) + \csc \left ( \frac{5\pi }{6} \right ) + 3\tan ^{2}\left ( \frac{\pi }{6} \right ) = 6$

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

$\cot \frac{\pi}{6} = \sqrt{3}\\ \\ cosec\frac{5\pi}{6} = cosec\left ( \pi - \frac{\pi}{6} \right )=cosec\frac{\pi}{6} = 2\\ \\ \tan\frac{\pi}{6}= \frac{1}{\sqrt{3}}$

$\cot^{2}\frac{\pi}{6} + cosec\frac{5\pi}{6} +3\tan^{2}\frac{\pi}{6} = \left ( \sqrt(3) \right )^{2} + 2 + 3\times\left ( \frac{1}{\sqrt{3}} \right )^{2}\\ \\ \Rightarrow 3+2+1 = 6$
R.H.S.

Question:4 Prove that $\small 2\sin ^{2}\left ( \frac{3\pi }{4} \right ) + 2\cos ^{2}\left ( \frac{\pi }{4} \right ) + 2\sec ^{2}\left ( \frac{\pi }{3} \right ) = 10$

Answer:

$\sin \frac{3\pi}{4} = \sin\left ( \pi-\frac{\pi}{4} \right ) = \sin \frac{\pi}{4}= \frac{1}{\sqrt{2}}\\ \\ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\\ \\ \sec\frac{\pi}{3}= 2$
Using the above values

$2\sin^{2}\frac{3\pi}{4} +2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3} = 2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}\\ \\ \Rightarrow 1+1+8=10$
R.H.S.

Question:5(i) Find the value of $\small (i) \sin 75\degree$

Answer:

$\sin 75\degree = \sin(45\degree + 30\degree)$
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

$\sin 75\degree = \sin(45\degree + 30\degree) = \sin45\degree\cos30\degree + \cos45\degree\sin30\degree\\ \\ \Rightarrow \frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\times\frac{1}{2}\\ \\ \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Question:5(ii) Find the value of
$\small (ii) \tan 15\degree$

Answer:

$\tan 15\degree = \tan (45\degree - 30\degree)$
We know that,

$\left [ \tan(x-y)= \frac{\tan x - \tan y}{1+\tan x\tan y} \right ]$
By using this we can write

$\tan (45\degree - 30\degree)= \frac{\tan 45\degree - tan30\degree}{1+\tan45\degree\tan30\degree}\\ \\ \Rightarrow \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3} -1\right )}=\frac{3+1-2\sqrt{3}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}}\\ \\ \Rightarrow \frac {4-2\sqrt{3}}{3-1}=\frac{2\left ( 2-\sqrt{3} \right )}{2}= 2-\sqrt{3}$

Question:6 Prove the following: $\small \cos \left ( \frac{\pi }{4}-x \right )\cos \left ( \frac{\pi }{4}-y \right ) - \sin \left ( \frac{\pi }{4} -x\right )\sin \left ( \frac{\pi }{4}-y \right ) =\sin (x+y)$

Answer:

$\cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) - \sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right )$

Multiply and divide by 2 both cos and sin functions
We get,

$\frac{1}{2}\left [2 \cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) \right ] + \frac{1}{2}\left [- 2\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right ) \right ]$

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities

In our question A = $\left (\frac{\pi}{4}-x \right )$

B = $\left (\frac{\pi}{4}-y \right )$
So,

$\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} + \cos \left \{ \left ( \frac{\pi}{4}-x \right) -\left ( \frac{\pi}{4}-y \right ) \right \} \right ] +\\ \\ \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} - \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$\Rightarrow 2 \times \frac{1}{2} \left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right )+\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$

As we know that

$(\cos \left ( \frac{\pi}{2} - A \right ) = \sin A)$
By using this

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$ $=\sin(x+y)$

R.H.S

Question:7 Prove the following $\small \frac{\tan \left ( \frac{\pi }{4}+x \right )}{\tan \left ( \frac{\pi }{4} -x\right )} = \left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$

Answer:

As we know that

$(\tan (A +B ) = \frac {\tan A + \tan B}{1- \tan A\tan B})$ and $\tan (A-B) = \frac {\tan A - \tan B }{1+ \tan A \tan B}$

So, by using these identities

$\frac{\tan \left ( \frac{\pi}{4}+x \right )}{\tan \left ( \frac{\pi}{4}-x \right )} = \frac{\frac{\tan \frac {\pi}{4} + \tan x}{1- \tan \frac{\pi}{4}\tan x}} {\frac{\tan \frac {\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4}\tan x}} =\frac{ \frac {1+\tan x }{1- \tan x}} { \frac {1-\tan x }{1+ \tan x}} = \left ( \frac{1 + \tan x}{1 - \tan x} \right )^{2}$
R.H.S

Question:8 Prove the following $\small \frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos \left ( \frac{\pi }{2}+x \right )} = \cot ^{2} x$

Answer:

As we know that,
$\cos(\pi+x) = -\cos x$ , $\sin (\pi - x ) = \sin x$ , $\cos \left ( \frac{\pi}{2} + x\right ) = - \sin x$
and
$\cos (-x) = \cos x$

By using these our equation simplify to

$\frac{\cos x \times -\cos x}{sin x \times - \sin x} = \frac{- \cos^{2}x}{-\sin^{2}x} = \cot ^ {2}x$ $(\because \cot x = \frac {\cos x}{\sin x})$
R.H.S.

Question:9 Prove the following $\small \cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] = 1$

Answer:

We know that

$\cos \left ( \frac{3\pi}{2}+x \right ) = \sin x\\ \\ \cos (2\pi +x)= \cos x\\ \\ \cot\left ( \frac{3\pi}{2} -x\right ) = \tan x\\ \\ \cot (2\pi + x) = \cot x$

So, by using these our equation simplifies to

$\cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] \\=\sin x\cos x [\tan x + \cot x] = \sin x\cos x [\frac {\sin x}{\cos x} + \frac{\cos x}{\sin x}]\\ \\ \Rightarrow \sin x\cos x\left [ \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x } \right ] =\sin^{2}x+\cos^{2}x = 1$ R.H.S.

Question:10 Prove the following $\small \sin (n+1)x\sin(n+2)x + \cos(n+1)x\cos(n+2)x =\cos x$

Answer:

Multiply and divide by 2

$= \frac {2\sin(n+1)x \sin(n+2)x + 2\cos (n+1)x\cos(n+2)x}{2}$

Now by using identities

-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)

$\frac{\left \{ -\left (\cos(2n+3)x - \cos (-x) \right ) + \left ( \cos(2n+3) +\cos(-x) \right )\right \}}{2}\\ \\ \left ( \because \cos(-x) = \cos x \right )\\ \\ = \frac{2\cos x}{2} = \cos x$

R.H.S.

Question:11 Prove the following $\small \cos \left ( \frac{3\pi }{4}+x \right ) - \cos\left ( \frac{3\pi }{4} -x\right ) = -\sqrt{2} \sin x$

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

$\cos \left ( \frac {3\pi}{4}+x \right ) - \cos \left ( \frac {3\pi}{4}-x \right ) = -2\sin\frac{3\pi}{4}\sin x = -2\times \frac{1}{\sqrt{2}}\sin x\\ \\ = -\sqrt{2}\sin x$ R.H.S.

Question:12 Prove the following $\small \sin^{2}6x - \sin^{2}4x = \sin2x\sin10x$

Answer:

We know that
$a^{2} - b^{2} = (a+b)(a-b)$

So,
$\sin^{2}6x - \sin^{2}4x =(\sin6x + \sin4x)(\sin6x - \sin4x)$

Now, we know that

$\sin A + \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )\\ \\ \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx

$\Rightarrow \sin^{2}6x - \sin^{2}4x = (2\cos5x\sin5x)(2\sin x\cos x)$

Now,

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0

hence
$\sin^{2}6x-\sin^{2}4x = \sin2x\sin10x$

Question:13 Prove the following $\small \cos^{2}2x - \cos^{2}6x = \sin4x\sin8x$

Answer:

As we know that

$a^{2}-b^{2} =(a-b)(a+b)$

$\therefore \cos^{2}2x -\cos^{2}6x = (\cos2x-\cos6x)(\cos2x+\cos6x)$
Now
$\cos A - \cos B = -2\sin\left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )\\ \\ \cos A + \cos B = 2\cos\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $\because$ sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes

R.H.S.

Question:14 Prove the following $\small \sin2x +2\sin4x + \sin6x = 4\cos^{2}x\sin4x$

Answer:

We know that
$\sin A+ \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $\because$ cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( $\because \cos2x = 2\cos^{2}x - 1$ )
=2sin4x( $2\cos^{2}x - 1$ +1 )
=2sin4x( $2\cos^{2}x$ )
= $4\sin4x\cos^{2}x$
R.H.S.

Question:15 Prove the following $\small \cot4x(\sin5x + \sin3x) = \cot x(\sin5x - \sin3x)$

Answer:

We know that
$\sin x + \sin y = 2\sin\left ( \frac{x+y}{2} \right )\cos\left (\frac{x-y}{2} \right )$
By using this , we get

sin5x + sin3x = 2sin4xcosx

$\frac{\cos4x}{\sin4x}\left ( 2\sin4x\cos x \right ) = 2\cos4x\cos x\\ \\$

now nultiply and divide by sin x

$\\\ \\ \frac{2\cos4x\cos x \sin x}{\sin x } \ \ \ \ \ \ \ \ \ \ \\ \\ =\cot x (2\cos4x\sin x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{\cos x}{\ sin x} = \cot x \right )\\ \\$

Now we know that

$\\ 2\cos x\sin y = \sin(x+y) - \sin(x-y)\\ \\$

By using this our equation becomes

$\\ \\=\cot x (\sin5x - sin3x)\\$
R.H.S.

Question:16 Prove the following $\small \frac{\cos 9x - \cos 5x}{\sin17x - \sin3x} = -\frac{\sin2x}{\cos10x}$

Answer:

As we know that

$\\ \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \cos 9x - \cos 5x = -2\sin 7x \sin2x \\ \\ \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \sin 17x - \sin 3x = 2\cos10x \sin7x\\ \\ \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} =\frac{-2\sin 7x \sin2x}{2\cos10x \sin7x} = -\frac{\sin 2x}{\cos10x}$
R.H.S.

Question:17 Prove the following $\small \frac{\sin5x + \sin3x}{\cos5x + \cos3x} = \tan4x$

Answer:

We know that

$\\ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\\and\\ \\ \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\$

We use these identities

$\\ \sin5x + \sin3x = 2\sin4x\cos x\\ \cos5 x + \cos 3x = 2\cos4x\cos x \\ \\ \frac{\sin5x + \sin3x}{\cos5 x + \cos 3x} = \frac{ 2\sin4x\cos x}{2\cos4x\cos x} = \frac{\sin4x}{\cos 4x} = \tan 4x$
R.H.S.

Question:18 Prove the following $\small \frac{\sin x - \sin y}{\cos x+\cos y} = \tan \frac{(x-y)}{2}$

Answer:

We know that
$\sin x - \sin y = 2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}\\and \\ \\ \cos x +\cos y = 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}\\$

We use these identities

$\\ We \ use \ these \ identities\\ \\ \frac{\sin x - \sin y}{\cos x +\cos y} =\frac{2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}}{ 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan \frac{x-y}{2}$

R.H.S.

Question:19 Prove the following $\small \frac{\sin x + \sin 3x}{\cos x + \cos3x} = \tan2x$

Answer:

We know that

$\\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\and\\ \\ \cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\ \\ We \ use \ these \ equations \\ \\ \sin x + \sin3x = 2\sin2x\cos(-x) = 2\sin2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\ \\ \cos x + \cos3x = 2\cos2x\cos(-x) =2\cos2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\ \\ \frac{\sin x + \sin3x}{\cos x + \cos3x} = \frac {2\sin2x\cos x}{2\cos2x\cos x}= \frac{\sin2x}{\cos2x} = \tan2x$ R.H.S.

Question:20 Prove the following $\small \frac{\sin x - \sin 3x}{\sin^{2}x-\cos^{2}x} = 2\sin x$

Answer:
We know that

$\sin3x = 3\sin x - 4\sin^{3}x \ \ \ , \ \ \cos^{2}-\sin^{2}x = \cos2x \\and \\ \cos2x = 1 - 2\sin^{2}x \\$

We use these identities

$\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \cos^{2}x- \sin^{2} = \cos2x\\ \cos2x = 1 - 2\sin^{2}x$ $\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ (\cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^(2)x - 1} = 2\sin x$ $\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^{2}x - 1} = 2\sin x$
R.H.S.

Question:21 Prove the following $\small \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Answer:

We know that

$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\ and \\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
We use these identities

$\frac{(\cos4x + \cos2x) + \cos3x}{(\sin4x+\sin2x)+\sin3x} = \frac{2\cos3x\cos x + \cos3x}{2\sin3x\cos x+\sin3x} = \frac{2\cos3x(1+\cos x)}{2\sin3x(1+\cos x)}\\ \ \ \\ \ \ \ \ \ \ \ = cot 3x$

=RHS

Question:22 prove the following $\small \cot x \cot2x - \cot2x\cot3x - \cot3x\cot x =1$

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

$cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$
So,
$cotx\ cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(cot2x+cotx)$
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.

Question:23 Prove that $\small \tan4x = \frac{4\tan x(1-\tan^{2}x)}{1-6 \tan^{2}x+\tan^{4}x}$

Answer:

We know that

$tan2A=\frac{2\tan A}{1 - \tan^{2}A}$

and we can write tan 4x = tan 2(2x)
So, $tan4x=\frac{2\tan 2x}{1 - \tan^{2}2x}$ = $\frac{2( \frac{2\tan x}{1 - \tan^{2}x})}{1 - (\frac{2\tan x}{1 - \tan^{2}x})^{2}}$

= $\frac{2 (2\tan x)(1 - \tan^{2}x)}{(1-\tan x)^{2} - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{(1)^{2}+(\tan^{2} x)^{2} - 2 \tan^{2} x - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{1^{2}+\tan^{4} x - 6 \tan^{2} x }$ = R.H.S.

Question:24 Prove the following $\small \cos4x = 1 - 8\sin^{2}x\cos^{2}x$

Answer:

We know that
$cos2x=1-2\sin^{2}x$
We use this in our problem
cos 4x = cos 2(2x)
= $1-2\sin^{2}2x$
= $1-2(2\sin x \cos x)^{2}$ $(\because \sin2x = 2\sin x \cos x)$
= $1-8\sin^{2}x\cos^{2}x$ = R.H.S.

Question:25 Prove the following $\small \cos6x = 32\cos^{6}x -48\cos^{4}x + 18\cos^{2}x-1$

Answer:

We know that
cos 3x = 4 $\cos^{3}x$ - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 $\cos^{3}2x$ - 3 cos 2x
= $4(2\cos^{2}x - 1)^{3}$ - $3(2\cos^{2}x - 1)$ $(\because \cos 2x = 2\cos^{2}x - 1)$
= $4[(2cos^{2}x)^{3} -(1)^{3} -3(2cos^{2}x)^{2}(1) + 3(2cos^{2}x)(1)^{2}]$ $-6\cos^{2}x + 3$ $(\because (a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2})$
= 32 $cos^{6}x$ - 4 - 48 $cos^{4}x$ + 24 $cos^{2}x$ - $6\cos^{2}x + 3$
= 32 $cos^{6}x$ - 48 $cos^{4}x$ + 18 $cos^{2}x$ - 1 = R.H.S.

Benefits of NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3

The NCERT book class 11th maths chapter 3 exercise once completed properly, can help in other subjects also.
Exercise 3.3 Class 11 Maths needs an analytical approach which can be developed by practice only.
Class 11 maths chapter 3 exercise 3.3 solutions are prepared by referring various good sources. Hence students can rely on these without any doubt.

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Frequently Asked Questions (FAQs)

1. Is it very tough to solve some of the questions from this exercise?

Yes, in that case one can refer to the solutions provided here.

2. Can there be different methods to solve questions ?

Yes, in proof related questions, more than one method can be used.

3. Can I get some marks if I am not able to complete the solutions but have done a few steps in the CBSE exam ?

Yes, there are marks for the steps in the CBSE exams

4. In how much time can one master exercise 3.3 maths class 11 ?

This exercise can take 5 to 6 hours if done properly in step by step manner.

5. How many questions are there in exercise 3.3 class 11 maths ?

There are 25 questions in this exercise, mostly proof related questions.

6. Can I skip some of the questions from this chapter for the CBSE exam ?

Not recommended. Can do if there is paucity of time .

7. Can one score 100 percent marks in this chapter questions ?

Yes, as most of the questions are simple and Maths is a very scoring subject.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 3.3 Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

More About NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3

Benefits of NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3

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